Search

structure diagram of hypophosphate anion

Input interpretation

hypophosphate anion | structure diagram
hypophosphate anion | structure diagram

Result

Draw the Lewis structure of hypophosphate anion. Start by drawing the overall structure of the molecule, ignoring potential double and triple bonds:  Count the total valence electrons of the oxygen (n_O, val = 6) and phosphorus (n_P, val = 5) atoms, including the net charge: 6 n_O, val + 2 n_P, val - n_charge = 50 Calculate the number of electrons needed to completely fill the valence shells for oxygen (n_O, full = 8) and phosphorus (n_P, full = 8): 6 n_O, full + 2 n_P, full = 64 Subtracting these two numbers shows that 64 - 50 = 14 bonding electrons are needed. Each bond has two electrons, so we expect that the above diagram has all the necessary bonds. However, to minimize formal charge oxygen wants 2 bonds. Identify the atoms that want additional bonds and the number of electrons remaining on each atom. The net charge has been given to the most electronegative atom, oxygen, in 4 places:  In order to minimize their formal charge, atoms with large electronegativities can force atoms with smaller electronegativities on period 3 or higher to expand their valence shells. The electronegativities of the atoms are 2.19 (phosphorus) and 3.44 (oxygen). Because the electronegativity of phosphorus is smaller than the electronegativity of oxygen and the electronegativity of {} is smaller than the electronegativity of {}, expand the valence shell of phosphorus to 5 bonds (the maximum number of bonds it can accomodate) in 2 places and expand the valence shell of {} to {} bonds. Therefore we add a total of 2 bonds to the diagram, noting the formal charges of the atoms. Double bonding phosphorus to the other highlighted oxygen atoms would result in an equivalent molecule: Answer: |   |
Draw the Lewis structure of hypophosphate anion. Start by drawing the overall structure of the molecule, ignoring potential double and triple bonds: Count the total valence electrons of the oxygen (n_O, val = 6) and phosphorus (n_P, val = 5) atoms, including the net charge: 6 n_O, val + 2 n_P, val - n_charge = 50 Calculate the number of electrons needed to completely fill the valence shells for oxygen (n_O, full = 8) and phosphorus (n_P, full = 8): 6 n_O, full + 2 n_P, full = 64 Subtracting these two numbers shows that 64 - 50 = 14 bonding electrons are needed. Each bond has two electrons, so we expect that the above diagram has all the necessary bonds. However, to minimize formal charge oxygen wants 2 bonds. Identify the atoms that want additional bonds and the number of electrons remaining on each atom. The net charge has been given to the most electronegative atom, oxygen, in 4 places: In order to minimize their formal charge, atoms with large electronegativities can force atoms with smaller electronegativities on period 3 or higher to expand their valence shells. The electronegativities of the atoms are 2.19 (phosphorus) and 3.44 (oxygen). Because the electronegativity of phosphorus is smaller than the electronegativity of oxygen and the electronegativity of {} is smaller than the electronegativity of {}, expand the valence shell of phosphorus to 5 bonds (the maximum number of bonds it can accomodate) in 2 places and expand the valence shell of {} to {} bonds. Therefore we add a total of 2 bonds to the diagram, noting the formal charges of the atoms. Double bonding phosphorus to the other highlighted oxygen atoms would result in an equivalent molecule: Answer: | |